Determining Active and Passive Pressures for Cohesion and

Cohesionless Soils Using Variation for a Smooth Wall

by Farid A. Chouery,1 P.E., S.E.

©2012 by Farid A. Chouery all rights reserved

Abstract

Variational method is used to determine active and passive forces for a smooth wall with

a cohesionless and cohesive back fill. The resulting slip surface shows that the extremum

of the force occurs when the slip surface is the Coulomb line for a level backfill. The

analysis shows that to achieve the Coulomb line the internal shear in the slices must be

zero. For a sloped backfill with cohesion the resulting slip surface is not a line. Adding

internal shear in the slices for the passive condition with a sloped backfill with a

cohesiveless material is examined and compared with log-spiral method. For special

boundary conditions, where the slip surface cannot be a line, the forces, the slip surface,

the pressure on the wall and the location of the resultant on the wall are obtained for

active and passive conditions. The method is adequate and will always give the derived

forces and slip surfaces.

Introduction

The problem of active and passive earth pressure for a smooth wall has been solved by

Coulomb theory (1776) [4], Rankine theory (1857) [8], and log-spiral theory after

Terzaghi (1941 & 1943) [11,12]. These theories are presented in most soil mechanics’

1 Structural, Electrical and Foundation Engineer, FAC Systems Inc., 6738 19th Ave. NW, Seattle, WA

2

texts [12,13]. Each of these methods involves assuming a line slip surface as an

approximation of a more complex surface as observed in both model tests and field

observations. The reason in the differences between these slip surface methods and the

laboratory is in the laboratory the observed slip surface is that of a slope stability problem

and not of a wall pressure problem. In this paper, a more general solution to the problem

is developed using variational methods using slices. In variational method, a general form

of failure surface is derived, and then "varied" until the extreme condition of pressure on

a wall is found. In a similar approach, variational method has been used to find the factor

of safety in slope stability: Baker and Garber (1978) [1] and numerical methods by

Leshechinsky (1990) [6]. The pressure discussed in this paper is for homogenous soil for

a smooth wall. It can be readily extended to multilayer soil. However, further work is

required. The limitation of the variational method is that it assumes a continuous function

for the slip surface. This can be a problem for a heterogynous material as seen in Fig. 1.

The method fails and

FIG. 1 - Heterogynous Material with Slip Surfaces

Slip Surfaces

3

revisions are needed, possibly piecewise functions are needed. Note: piesewise continous

function will be needed for multilayer homogenous soil. The general methodology has

been classically to find the extremum for the earth pressure by finding a slip surface

where a potentially soil failure would occur for extremum pressure. Thus, the design has

captured the worst condition for the force on the wall. Historically, this method has been

used satisfactorily provided the material properties have been reached. It is impossible in

practice to test for all the soil friction values and cohesion through borings logs, thus

engineering judgment has to be made for these values. In order to use the method

proposed properly and interpret the results, care has to be taken in selecting the material

properties.

With the variational method, [15], one selects arbitrary admissible slip surfaces and

determines the forces acting on the boundary of the earth mass. The definitive slip surface

is one that yields an extremum value for the earth pressure. In this paper, variational

methods are used to determine the slip surface, based on some practical assumptions. The

solution shows that the Coulomb wedge is a particular case of the general solution

presented here. It is believed that the application of variational methods to earth pressure

problems is a practical advancement to understanding of retaining wall problems. Five

different analyses will be done to avoid giving one cumbersome general equation and

give a closed form solution for the special cases. In analysis 1 a closed form solution is

reached, it is assumed a level backfill for cohesive and cohesiveless material, which is

4

useful for most site conditions. Analysis 2 is for a sloped backfill and slanted wall for

cohesiveless material, which does not include cohesion to give a closed form solution.

Analysis 3 is repeating analysis 2 with cohesion where the pressure is not given in a

closed form but can be obtained numerically. In analysis 4 a closed form solution is

reached for passive earth pressure condition for sloped backfill with cohesiveless material

to include internal vertical shear from the slope backfill. Analysis 5 is a repeat of Analysis

4 with cohesion.

Analysis 1

In this analysis, it is desired not to restrict the slip surface to a line and to find a function

y(x) that would extremize E, as shown in Fig. 2(a). It is assumed that the wall moves

dE

dW

y(x)

dx0

(0,y )

x

y

dx

T

EE

TdW

0

n

n

n+1

n+1

α

α

α

α−φ

φ

Proposed Slip Surface A Slice

FIG. 2(a) FIG. 2(b)

dQ

( )x y0 1,

FIG. 2 – Slip Surface for Level Backfill

5

sufficiently such that the friction between the wedges equal 0. Thus, Tn+1 = Tn , as in Fig.

2(b), for all the wedges. Since the wall has no friction, Ti = 0 for all the wedges. This

assumption is similar to the assumptions used in stability analysis by the method of slices

(Bishop (1955) [2]) and in analysis of passive earth pressures for a wall with friction by

Shields and Tolunay (1973) [10]. If the Ti 's are to be included in the derivations, K0 can

be determined using incipient shear method. Thus, each slice is in equilibrium, so the

overturning moment remains in balance and not of concern.

Thus, the force dE can be written as

[ ])tan(sincos)tan(1 φαααφα −+−−==−+ cdsdWdEEE nn ................................ (1)

Replacing dW by γydx and ds by dx/cosα and integrating yields

[ ]{ }∫ −+−−=0

0)tan(tan1)tan(

x

dxcyE φααγφα ...................................................... (2)

or

( )dxcyE

x

∫

+

−+−

+−

=0

0 tantan1

tantantan1

tantan1

tantan

φαφαα

γφαφα

................ ...................... (3)

Now tanα = − = ′dy

dxy ................................................................................................ (4)

Substituting Eq. 4 in Eq. 3 yields

( )dx

y

yycy

y

yE

x

∫

′−−′−′

−−′−−′−

=0

0 tan1

tan1

tan1

tan

φφ

γφφ

.................................................... (5)

6

Eq. 5 can written as:

( )dx

yycy

yE

x

∫

′−

+′−−−

′−+

−=0

0 22 tan1sin

1 cot

sin

11

tan1

cottancot

φφφ

φγ

φφφ

φ

…………………………………… (6)

It is necessary to find the extremum of E in Eq. 6 for the boundary condition:

(i) Given (0, y0) and (x0, y1), find the slip surface that passes through these points.

(ii) Given (0, y0) and P0 located somewhere on a given line x = x0 or a line y = y1,

find the slip surface.

Therefore Euler’s equation [15] applies where

( )

0 and

tan1sin

1 cot

sin

11

tan1

cottancot

22

=

′

ℜ−

ℜ

′−

+′−−−

′−+

−=ℜ

ydx

d

y

yycy

y

∂∂

∂∂

φφφ

φγ

φφφ

φ

.............. (7)

which can be written as ℜ − ′ℜ

′=y

yh

∂∂ 0 ............................................................... (8)

where ℜ does not involve x explicitly, and h0 is a constant.

Substituting in Eq. 8 yields

7

( )

( ) ( ) 0222

2

22

tan1sin

tan

tan1sin

1cot

tan1costan1

cottancot

hy

y

yc

yy

y

y

=

′−

′−

′−+−−

′−

′+

′−+

−

φφφ

φφφ

γφφφ

φφφ

....................... (9)

Eq. 9 reduces to the following:

( )( ) ( )2

02 tan1tan1tan2 φφγφ yhcyyy ′−=+−′+′ ...................................................... (10)

or

( )( ) ( )2

0 tan1)tan(cos/1tancos/1tan φφγφφφφ yhcyyy ′−=+−+′++′ ................... (11)

Note Eq. 10 is a parabola in ′y . From Eq. 11: h y0 0 1≥ ′ ≤ − − for tan / cosφ φ or

′ ≥ − +y tan / cosφ φ1 , and h y0 0 1 1< − − ≤ ′ ≤ − + for tan / cos tan / cosφ φ φ φ . From

trigonometric identity

( )( )tan tan

/ cos /

sin /

sin

costan

cos

π φ π φ π φπ φ

φφ

φφ4 2

2

2

1 2

2

1 1±

=

±

=

− ±

±=

±= ± + ......... (12)

From Eq. 11 and Eq. 12, ( )h0 0 4 2 4 2≥ ≥ + ≤ − − for or α π φ α π φ/ / / / , and

( )h0 0 4 2 4 2< − − ≤ ≤ + for π φ α π φ/ / / / . Taking α to be positive yields the

following bound on h0 :

2/4/0for 0 and , 2/4/for 0 00 φπαφπα +≤≤<+≥≥ hh ......Active............. (13)

8

For the passive condition, replacing φ by −φ and c by –c in Eq. 11 yields

2/4/0for 0 and , 2/4/for 0 00 φπαφπα −≤≤<−≥≥ hh ......Passive............ (14)

The slip surface can be derived by rewriting Eq. 10 to

( ) ( ) ( ) 0tan2tan 1122

1 =+−′++′− hyyhyyhy φφ ....................................................... (15)

where φγ

φγtan

and

tan1

01

hh

y

c

hh =

+= is a new constant. Rewriting Eq. 15 in

′ ′x y instead of yields (See Appendix I)

hc

y

cy

x

++

+±=′

φγ

φγφ

φ

tan

tan

cos

1tan …………………………………………………. (16)

If h = 0 in Eq. 16, then ′x becomes a constant and y(x) must follow the Coulomb wedge.

Since ′ = −x cotα and α π φ= +/ /4 2 for a Coulomb wedge it follows from Eq. 12 that

( ) ( )′ = − + = − − = −x cot / / tan / / tan / cosπ φ π φ φ φ4 2 4 2 1 ................................. (17)

9

In order to satisfy Eq. 17, the plus sign in Eq. 16 can be dropped, and the equations

become

hc

y

cy

x

++

+−=′

φγ

φγφ

φ

tan

tan

cos

1tan ..............Active........................................... (18a)

For the passive condition, replacing φ by −φ and c by –c

hc

y

cy

x

++

+−−=′

φγ

φγφ

φ

tan

tan

cos

1tan ..............Passive.......................................... (18b)

Integrating Eq. 18a yields the slip surface equation

khc

yc

yhc

yh

cyh

cyyx

+

+

++

++

+−

++

+−=

φγφγφγ

φγφγφφ

tan2

tantan2ln

2

tantancos

1tan

2

2

...................................... (19)

where k is the constant of integration. Substituting the point at x = 0 y = y0 yields

10

khc

yc

yhc

yh

cyh

cyyk

+

+

++

++

+−

++

++−=

φγφγφγ

φγφγφφ

tan2

tantan2ln

2

tantancos

1tan

00

2

0

0

2

00

................................ (20)

and the active slip surface, Eq. 19, becomes

+

++

++

+

+

++

++

+

−

++

+−

++

+−−=

hc

yc

yhc

y

hc

yc

yhc

yh

cyh

cy

cyh

cyyyx

φγφγφγ

φγφγφγ

φγφγφγφγφφ

tan2

tantan2

tan2

tantan2

ln2

tantantantancos

1tan)(

00

2

0

2

0

2

0

2

0

…………………….. (21a)

For passive it becomes

+

++

++

+

+

++

++

+

−

++

+−

++

+−−−=

hc

yc

yhc

y

hc

yc

yhc

yh

cyh

cy

cyh

cyyyx

φγφγφγ

φγφγφγ

φγφγφγφγφφ

tan2

tantan2

tan2

tantan2

ln2

tantantantancos

1tan)(

00

2

0

2

0

2

0

2

0

……………...………………..(21b)

To find the active force E, Eq. 5 can be rewritten in terms of ′ ′x y instead of ,

11

dy dx instead of , and the interval [ ] instead of [ ] , where y x y0 0 10 0, , is taken to be 0 in

Fig. 2(a). Thus

( )dy

x

xcyx

x

xE

y∫

′−

′++′

′−

′+=

02

0 tan

1

tan

tan1

φγ

φφ

.......................................................................................... (22)

Substituting Eq. 18a in Eq. 22 and rearranging yields

dy

cyh

cy

hc

y

c

dy

cyh

cy

hc

y

yE

y

y

∫

∫

++

+

+

+

−−

++

+

+

+

−+=

0

0

0 2

0 22

2

tantan

tan2

cos

1tan2

tantan

tan2

cos

tan

cos

1tan

φγφγ

φγφ

φ

φγφγ

φγφφ

φφγ

......................................................... (23)

Thus the mathematical pressure q ydE

dy( ) = , from each slice to another, becomes

++

+

+

+

−−

++

+

+

+

−+==

φγφγ

φγφ

φ

φγφγ

φγφφ

φφγ

tantan

tan2

cos

1tan2

tantan

tan2

cos

tan

cos

1tan)(

2

22

2

cyh

cy

hc

y

c

cyh

cy

hc

y

ydy

dEyq

12

..................................................... (24)

It is desired to investigate the extremum of q(y) with respect to h, thus

0 yields 0 == hdh

dq ..................................................................................................... (25)

Thus the extremum occurs at the Coulomb wedge for passive and active. Hence it has

been shown that the Coulomb wedge is the wedge for a smooth wall that moves

sufficiently such that Tn+1 = Tn . It is noted that all of Ti = 0 since the wall is smooth. For

other prescribed conditions (i) and (ii) above it is necessary to use the slip surface of Eqs.

21a-b. Integrating Eq. 23 yields the active force

φγφγφγ

φγφγφγ

φφ

γ

φγφγφγφγφγφγφφ

γ

φφ

φγ

tan

2

tantan2

tan2

tantan2

ln4cos

tan

tantan2tantantan2tancos

tan

tan2cos

1tan

2

2

00

2

02

2

0

2

00

02

220

ch

ch

c

cyh

cyh

cy

h

ch

chccyh

cy

hcy

cyy

E

++

+

+++

++

+

−

+

−−

++

+

−+−

+

+=

……………………….…............. (26a)

and the passive force

13

φγφγφγ

φγφγφγ

φφ

γ

φγφγφγφγφγφγφφ

γ

φφ

φγ

tan

2

tantan2

tan2

tantan2

ln4cos

tan

tantan2tantantan2tancos

tan

tan2cos

1tan

2

2

00

2

02

2

0

2

00

02

220

ch

ch

c

cyh

cyh

cy

h

ch

chccyh

cy

hcy

cyy

E

++

+

+++

++

+

+

+

−−

++

+

−++

+

+=

……………………………............ (26b)

Note: that the right hand term involving h in Eqs. 26a-b has a minimum at h=0. It is

desired to show that for φγ tan

ch −< the slip surface does not reach the top of ground at

the point (x0,0). It has been shown in Eqs. 13 and 14 that if h0 < 0 (or h < 0), then

α π φ α π φ≤ + ≤ −/ / / /4 2 4 2 for active, and for passive. Thus a slip surface with

φγ tan

ch −< is further than the triangle Coulomb wedge as shown in Fig. 3.

x0

y1

CoulombWedge

45+φ/2

Slip Surface for h < 0

Slip Surface

FIG 3 – Slip Surface for Level Backfill with Constraint h < 0

14

Furthermore, Eq. 18a-b shows that ′ = ∞ ′ =x y or 0 at some value φγ tan

chy −−= .

Since φγ tan

ch −< , the y value must be positive and cannot be 0. Thus y is below

ground. Also, for φγ tan

chy −−< the term h

cy ++

φγ tan is not defined in Eq. 18a-b.

Thus for φγ tan

ch −< Eqs. 26a-b are not applicable.

Eq. 22 must be integrated from y0 to y1 instead of y0 to 0, where the point (x0, y1) is a

point underground. This integration yields

+++

++

+

+++

++

+

−

++

+

−+−

++

+

−+−

−+

+

−=

φγφγφγ

φγφγφγ

φφ

γ

φγφγφγ

φγφγφγφφ

γ

φφ

φγ

tan2

tantan2

tan2

tantan2

ln4cos

tan

tantan2tan

tantan2tancos

tan

tan)(2cos

1tan

22

11

2

1

00

2

02

1

2

11

0

2

00

102

221

20

cyh

cyh

cy

cyh

cyh

cy

h

cyh

cy

hcy

cyh

cy

hcy

yycyy

E

..................................Active.................................... (27a)

15

+++

++

+

+++

++

+

+

++

+

−+−

++

+

−++

−+

+

−=

φγφγφγ

φγφγφγ

φφ

γ

φγφγφγ

φγφγφγφφ

γ

φφ

φγ

tan2

tantan2

tan2

tantan2

ln4cos

tan

tantan2tan

tantan2tancos

tan

tan)(2cos

1tan

22

11

2

1

00

2

02

1

2

11

0

2

00

102

221

20

cyh

cyh

cy

cyh

cyh

cy

h

cyh

cy

hcy

cyh

cy

hcy

yycyy

E

..................................Passive.................................. (27b)

Thus it can be easily concluded that h = 0 is the only minimum of the h terms in Eqs.

26a-b. Hence E has a maximum at h = 0 for active and a minimum at h = 0 for passive.

This shows that the Coulomb wedge gives the extremum for E as it has been shown in

Eq. 25.

Condition (i)

It is seen that Eqs. 18a-b , 21a-b , 26a-b , and 27a-b are useful for condition (i) when the

slip surface must pass through a point and physically is not represented by the Coulomb

wedge. The following examples have this situation:

Example 1 Consider Fig. 4, where the passive pressure is to be calculated for a slip

16

x0

y0

y2

45−φ/2

(0,y0)

(x0,0)

Slip Surface of Eq. 31-b

Basement Wall

FIG. 4 – Slip Surface for Passive with and Existing Basement Constrained Example 1

surface that is controlled by the presence of a basement wall. The Coulomb wedge cannot

physically go through the wall. Thus the general slip surface is applicable.

In this situation y0, x0 are known. Thus, substituting their values in Eq. 21b yields

x y y hyh h

y hy y h0 0 0

20

02

0 0

1

2 2 2= + + +

+ + +

tan

coslnφ

φ ................................ (28)

The value h can be obtained numerically from Eq. 28, and E can be calculated from Eq.

26b. Thus, if γ = 120 pcf (1.93 g/cm3), φ = 30 degrees, c = 0, y0 = 10 ft (3.05 m), and x0 =

10 ft (3.05 m), then from Eq. 28, h = 27.3318 ft (8.3307 m). Taking λ = h y/ 0 = 2.73321

in Eq. 26b gives E = 21,455 plf (31.97 g/m). If a straight line is used from (0, y0) to (x0,

0), then α = 45 degrees. Evaluating the Coulomb wedge y = y0 = 10 ft (3.05 m) gives E =

22,392 plf (33.37 g/m). Note that Eq. 26b gives the extremum with 4.4% difference over

the straight line. Note: If treated as the full wedge with Kp = 3 where y = 5.77ft (1.76 m)

is triangular pressure and 4.33 ft (1.29 m) is pressure with surcharge then E = 18,000 pfl

(26.82 g/m). In this situation it is assumed the wall is rigid thus the slip surface is a curve

and E = 21,455 plf (31.97 g/m).

17

Example 2 Consider Fig. 5, where a slip surface passing through (0, y0) to (x0, y1) must

y0=20

x0=5

y1=5

line 2line 1

71.565 deg

(x0,y1)

(0,y0)

FIG. 5 – Slip Surface for Active with and Existing Basement Constrained Example 2

be analyzed for an active condition. γ = 120 pcf (1.93 g/cm3), and φ = 30 degrees, c = 0.

From Eq. 21a substituting y0 = 20 ft (6.1 m), x = x0 = 5 ft (1.52 m), y = y1 = 5 ft (1.52 m),

and calculating numerically h = 6.88284 ft (2.0979 m). Substituting in Eq. 27a gives E =

6,740 plf (10.05 g/m). If comparing with a straight line, line 1, from (0, y0) to (x0, y1), it

gives E = 6,651 plf (9.91 g/m), where the Coulomb wedge was used with W =

.5(5+20)(5)(120)=7,500 plf (11.18 g/m), and α = 71.565 degrees. Thus, a 1.3% difference

over the derived is obtained. To find Emax further analysis must be done for line 2 and the

result must be compared with line 1.

Example 3 In the case of stability analysis for tieback wall, as shown in Fig. 6(a) , the

18

ψ

ξ

ψ

α

Q

WT

Pa

y0

x0

y2

paW

y1

Pa

pa

T

Q

E

W

Anchor

Soldier Pile

Tieback WallFIG. 6-a

Force DiagramFIG. 6-b

ξ

FIG. 6 - Slip Surface to Determine Safety Factor of a one Tieback Wall

slip surface is shown to pass through the point in the middle of the anchor, see reference

[3,7,9,16]. In this case the stability factor Tmax/Tdesign is to be calculated as in reference

[16]. From the force diagram, Fig. 6(b), the summation of horizontal and vertical forces

are

p T Q Pa a+ + =cos sinξ ψ ........................................................................................ (29)

T Q Wsin cosξ ψ+ = ................................................................................................ (30)

Substituting Q from Eq. 30 in Eq. 29 yields

19

TP W pa a=

− −

−

tan

cos sin tan

ψξ ξ ψ

............................................................................................... (31)

From Eq. 31 T = Tmax can be calculated where W Etanψ = of Eq. 27a, and the constant h

can be found from Eq. 21a for a given (x0, y1). W can be calculated numerically from

integrating the right hand side of Eq. 21a from y1 to y0

W y x f y dyy

y

= + ∫γ γ1 01

0

( ) ........................................................................................... (32)

where f(y) is the right hand side of Eq. 21a. Thus, ψ =

−tan 1 E

W,

p ya = −

1

2 4 212 2γ

π φtan , and P ya = −

1

2 4 202 2γ

π φtan .

Taking for example y0 = 20 ft (6.1 m), γ = 120 pcf (1.93 g/cm3), φ = 30 degrees, c = 0, ξ =

20 degrees, Tdesign = 3,872 plf (5.77 g/m), Pa = .333(120)(20)(20)/2 = 8,000 plf (11.92

g/m), y2 = 6 ft (1.83 m), x0 = 15 ft (4.57 m), y1 = 6 + 15 tan(20) = 11.46 ft (3.49 m), pa =

.333(120)(11.46)(11.46)/2 = 2,626 plf (3.91 g/m), and substituting x = x0 = 15 ft (4.57

m), and y = y1 = 11.46 ft (3.49 m) in Eq. 21a gives h = -10.8507 ft (-3.3073 m).

Substituting in Eq. 27a gives E = 154.4 plf (0.23 g/m). Integrating Eq. 32 numerically

gives W = 26,802 plf (39.94 g/m). Thus, ψ = 0.3301 degrees, and T in Eq. 31 becomes

5,566 plf (8.3 g/m). The stability factor Tmax/Tdesign = 5,566/3,872 = 1.438. If using a

straight line method, where ψ in Eq. 31 becomes α−φ, α = 29.66 degrees, and W =

0.5γ(y0+y1)x0 = 28,314 plf (42.2 g/m), then T = 5,885 plf (8.77 g/m), and the stability

20

factor = 5,885/3,872 = 1.52. This gives 6% difference in the safety factor over the derived

method.

Condition (ii)

For condition (ii), where the slip surface must pass through a line at x0 or at y1 below x0 ,

it yields ∂∂ℜ

′=

=y

x x0

0, or from Eq. 7

( ) ( )

( )0

coscot

tan1

tan

0cottan1sin

tan

tan1cos

2

2

1

2222

1

=−′−

+

=+′−

−′−

−

γφφ

φφγ

φφφ

φφφ

γ

c

y

cy

or

cy

cy

y

....................................... (33)

Substituting Eq. 18a with y1 instead of y in Eq. 33 where ′ = ′x y1 / yields

0tantancos

1

tantan

2

11 =−

+−++

φγφγφφγφ

ccyh

cy ……….....active….... (34a)

0tantancos

1

tantan

2

11 =−

+−++−

φγφγφφγφ

ccyh

cy ……...passive……. (34b)

Thus

21

+

±

+=

φγφγφφ

φγφφ

tantancos

tan2

tantan2tan

2

221

ch

cchy

.............................................................................................. (35)

At ∞→y& 0→x& then

φγφ

tantan 2

2

chy −= ...................only active ........................................................ (35a)

Substitute Eq. 35 in Eq. 21a and 21b for a given x0 and find h. If y1 is negative then y1 = 0

and h can be calculated from Eq. 21a and 21b for a given x0. If y1 is above y2 then y2

controls and find h from Eq. 21a and 21b. Note: y1 is zero only if h = 0 and c = 0. Thus, it

can be obtained from Eq. 34a is the following problem: Consider that the slip surface

must pass through a line at x = x0 and have y h12= tan φ below x0 and c = 0. This makes

′ = ′ = ∞x y0 or in Eq. 18a. This condition can occur in the real world as seen in

example 4 below, and E can be obtained from Eq. 27a. Note y h12= tan φ with c = 0 is

invalid for passive pressure, since Eq. 44b is not zero. For passive Eq. 45 is not valid

when replacing φ by −φ and c by –c and the slip surface must pass through the point (x0,

0) when there is a rigid obstacle otherwise it can be analysed as a surcharge problem as

seen in example 1.

For φγ tan

1

chy −−= , h < 0 and ′ = ′ = ∞y x0 or in Eq. 18a-b. This situation can happen

if the slip surface must pass through a line x = x0 where the slope must be zero at y1 below

x0 . Fig. 7 shows such an example of an passive condition with c = 0, where the concrete

22

x0

y0

y1 Concrete Mat

y' =0

FIG. 7 – Slip Surface for Passive with Mat Slab Constrained Special Condition

mat shown on top can sustain itself. Thus at the line x = x0 , ′ =y 0 due to deformation. In

this example the weight of the slab is taken to be the same as the weight of soil.

Substituting c = 0, h = -y1 and y = y1 in Eq. 21b yields

( )

−+−−−+−−=

1001

2

0

1101

2

0010

22ln

2cos

1tan

yyyyy

yyyyyyyx

φφ ............ (36)

and the passive force E can be obtained from Eq. 27b:

−+−+−

++

1

1001

2

02

101

2

01

0

22ln

42cos

tan

y

yyyyyyyyy

yy

φφγ

............................................................................................................. (37)

Example 4 Consider Fig. 8, where the active pressure is to be calculated for a slip surface

+

−=

φφγ

2

22

1

2

0

cos

1tan

22

yyE

23

x0

y0

y1

y3

(0,y0)

(x0,y1)

Basement Wall45+φ/2

FIG. 8 – Slip Surface for Active with a Neighboring Structure Example 4

that avoids the basement wall with c = 0. This problem is also similar to finding the

active pressure for a wall adjacent to rocks.

From Eq. 35

h y= 12cot φ .............................................................................................................. (38)

Substituting Eq. 38 in Eq. 21a yields

24

( )( )x

yz

zz

z z

z z

0

0

22 2

2 21

11

2

2 2

2 1 2= − − − + −

+ +

+ + +

tan

cos sincot

cotln

/ sin cot

cot cotφ

φ φφ

φ φ φ

φ φ

..................................................................................................... (39)

where z = y1/y0 . Thus, for a given x0 and y0 , z can be calculated numerically from Eq. 39.

Taking h y zy= =12

02cot cotφ φ , it can be substituted in Eq. 27a to give the active force.

Taking for example y0 = 20 ft (6.1 m), γ = 120 pcf (1.92 g/cm3), φ = 30 degrees, c = 0, x0

= 5 ft (1.52 m), gives z = 0.12475, y1 = 2.495 ft (76.05 cm), and h = 7.484 ft (228.14

cm). Substituting in Eq. 27a gives E = 6,777 plf (10.1 g/m). If a line is taken from (0, y0)

to (x0, 0), using the Coulomb wedge, it gives E = 6,530 plf (9.73 g/m), a 3.8% difference

over the derived. Now if a Coulomb method is used with y3 as shown in Fig. 8, the wedge

can be taken as the Coulomb line with a uniform surcharge γy3 . This gives E = 5,428 plf

(8.09 g/m), where the overburden y3 = 11.34 ft (3.46 m). Thus, the Coulomb wedge does

not give Emax and a difference of 25% is obtained over the derived.

For condition (ii), where the slip surface must pass through a line y = y1, ℜ in Eq. 7 can

be rewritten as

′−

′+−′−′

′−

′+−=ℜ

x

xxcxy

x

x

φφ

φφ

γtan

tan1

tan

tan1 .................................................................. (40)

If the Euler Eq. is used with Eq. 40, the same slip surface will be obtained. Thus, for a

slip surface passing through a line y = y1, it yields

25

∂∂ℜ

′=

=xy y1

0 ................................................................................................................. (41)

Executing Eq. 41 on Eq. 40 yields

( ) ( )0

)(tan

1tan2

)(tan

1tan2tan2

2

2

2

1 =′−

−′−′−

′−

−′−′

x

xxc

x

xxy

φφ

φφφ

γ .............................................. (42)

Thus,

′ = ± + = ±=x y y1

2 11

tan tan tancos

φ φφ

.................................................................. (43)

This forces h in Eq. 16 to be zero, or the Coulomb wedge is the solution for this

condition. This situation is similar to having a uniform surcharge at the line y = y1. This

confirms that the Coulomb wedge is the proper slip surface as described in Eq. 25 and in

the extremum of Eq. 26.

Analysis 2

For a sloped smooth wall with a sloped back fill with c = 0 (see Fig. 9(a)) the slip surface

26

x tan(β−θ) β−θ

β

α

φ

α−φ

α−φ

(0,y0)

(x0,y1)

dE

dWcosdWsin

TTi+1

i

θθ

θ θ

θdWcos

dE

dWsindQ

dQ

Proposed Slip SurfaceFIG. 9(a)

Force DiagramFIG. 9(b)

x

y

FIG. 9 - Slip Surface with a Sloped Backfill and Slanted Wall

for an active condition can be derived similarly. It is assumed that Ti = Ti+1 with T0 = 0.

From the force diagram, Fig. 9(b), it yields

( )[ ]dE dW= − +cos tan sinθ α φ θ ............................................................................. (44)

Thus,

( )[ ] ( )[ ]E y x dxx

= − + + −∫γ θ α φ θ β θcos tan sin tan0

0

............................................. (45)

where ( )[ ]dW y x dx= + −tan β θ . Eq. 45 can be rewritten as

27

E ku

uudx

x

= −′ +− ′

∫γ θ

ζφ

θ4 0 10

sintan

tancos .................................................................... (46)

where ( )k4 1= + −tan tanφ β θ , ζ φ β θ= − + , ( )[ ]u y x k= + −tan /β θ 4 , and

[ ]′ = ′ + −u y ktan( ) /β θ 4 . From variational method, and repeating the same analysis as in

analysis 1, it yields

dx

duk

u

u h= −

+tanφ 1 ............................................................................................... (47)

and the slip surface

khuhuuh

huukux +

+++−+−= 22ln2

tan 221φ ........................................ (48)

where ( )[ ] ( )k1 1= + −tan tan tan / tan tanφ ζ φ ζ θ , and k is the constant of integration.

Substituting Eq. 47 in Eq. 46 and rewriting yields

E k k ku h

u huudu

u

u

= −+

+

∫γ 4 3 2 2

2

1

0

.............................................................................. (49)

where

u y k0 0 4= / ,

( )[ ]u y x k1 1 0 4= + −tan /β θ ,

28

( )( )k2 1= + −cos tan tan tan tan tanθ φ ζ φ ζ θ , and

( )k3 1 2= − + +sin tan cos tan tanθ φ θ ζ φ .

Note E is maximum at h = 0, giving the slip surface to be the Coulomb wedge. Thus the

application of β and θ is the same as in Coulomb. Validation of the solution at h = 0 can

be checked numerically against US Army Corps Publication EM 1110-2-2502 Equation

3-13 and 3-19 as a different mathematical form of the same solution. Terzaghi (1943)

[12], Rosenfarb and Chen ( 1972) [9] show for high β and θ = 0 on a smooth wall, the

slip surface is not a line and produces lower Kp values. However, for Ka values the

differences were negligible. This means that the Ti's due to the ramp are not zero in

passive condition. This is possible for a high β > 0 influencing the slice contact shear at

the ramp. Equations derived for these conditions are shown in analyses 4 and 5 below.

Evaluating E in Eq. 49 yields

E k ku u

k k uh

u hu= −

− −

+

γ γ4 3

02

12

4 2 0 02

02 2 2

− −

+ +

+ + +

+ + +

uh

u huh u hu u h

u hu u h1 1

21

20

20 0

12

1 12 4

2 2

2 2ln

........................................................................................................ (50)

For passive replace φ by φ− and E becomes:

29

+

−+

−= 0

2

0024

2

1

2

034

222huu

hukk

uukkE γγ

+++

+++++

−−huhuu

huhuuhhuu

hu

11

2

1

00

2

02

1

2

11

22

22ln

42

........................................................................................................ (51)

For condition (i) described before, the constant h and k can be obtained by substituting

the two end points (0, u0) and (x0, u1) in the Eq. 48. For condition (ii): It is of merit to

obtain the general solution, for the undetermined points, for a given curve and not just for

a line. Some practical applications would be a buried vault, tunnel, tank, utility, etc. , that

would interfere with the coulomb line. Suppose the given curve is g0(x, y) = 0. By

substituting for y k u x= − −4 tan( )β θ in g0(x, y), the new function will be g(x, u) = 0. If

the end point (x0, u1) is obtained, then y1 can be obtained from y k u x1 4 1 0= − −tan( )β θ .

From reference [15] the following condition must be satisfied for condition (ii):

[ ]∂∂

∂∂

∂∂

∂∂

ℜ

′−

ℜ

+ ′=

=

==

= =

u

g

u

g

xu

g

ux x

u u

x x

x x u u

0

1

0

0 1

1

0 ................................................................................ (52)

where ℜ is the integrand of Eq. 46. Doing the mathematics of Eq. 52, and solving for ′u1

and using a relation for h from Eq. 47, yields

30

hR u k

R u kR u

u u=−

− + −+

−

−1

11

1 12

1 11

2

1 1

( ) cot

( )( )( tan tan )

(tan tan )

φ

φ θζ θ

.................................................. (53)

where

R u

g x u

x

g x u

u

x x

u u

( )

( , )

( , )10

1

==

=

∂∂

∂∂

................................................................................................. (54)

R(u1) is a function of u1 alone is because x0 can be determined from g(x0, u1) = 0. Now, k

in Eq. 48 is determined from the point (0, u0). Thus, substituting Eq. 52 in Eq. 48 and

replacing (x, y) by (x0, u1) gives an equation with u1 as the only unknown. Hence, u1 can

be determined (numerically).

The following are two important functions:

A line: [ ]y ax b k u a x b g x u= + → − − + − = = change to

4 0tan( ) ( , )β θ

from Eq 63

→ = −− +

R ua

k( )

tan( )1

4

β θ .......................................... (55)

A circle: ( ) ( ) ( ) ( tan( ) )x x y y r x x k u x y rc c c c− + − = → − + − − − − =2 2 2 24

2 2 0 to β θ

31

[ ]

to → =

−

− − −−

−R u

x x

k k u x y k

c

c

( )tan( )

tan( )1

0

4 4 1 0 4β θ

β θ ................ (56)

where x0 can be determined from the quadratic equation in g(x0, u1) = 0.

The following equations are the results for the condition where k4 = 0:

Eq. 45 yields:

Ek

uudx

x

= + +′

∫γ θ θ φsin cos cot 5

0

0

......................................................................... (57)

dx

du

h

u

k

k= − 6

52 ............................................................................................................ (58)

kuk

kuhx +−=

5

6

2ln ................................................................................................ (59)

E kh

u

k

ku du

u

u

= −

∫γ 5

26

2

540

1

......................................................................................... (60)

( )Ek

ku u k h

u

u= − −γ γ6

2

50

21

25

2 0

18ln ............................................................................ (61)

where ( )u y x= + −tan β θ , ′ = ′ + −u y tan( )β θ , u y0 0= , ( )u y x1 1 0= + −tan β θ ,

k52= cos / sinθ φ , ( )k6 = −cos / sinθ φ φ , and k is the constant of integration. For passive

replace φ by φ− . For Condition (ii):

hk R u

k

u

R u= − − −

1 1

26 1

5

2

1

1

( )

( ) ............................................................................... (62)

32

where R(u1) is of Eq. 54. To obtain the solution for a line or a circle, set k4 = 1 in Eqs. 55

and 56, and u1 can be determined numerically from Eq. 58.

For the passive condition replace φ by −φ in the above equations, starting with Eq. 44.

It is noteworthy that if the axis in Fig. 9(a) is rotated by −(β−θ), the new axis will be

( ) ( )x y x= − − + −sin cosβ θ β θ , ( ) ( )y y x= − + −cos sinβ θ β θ , and

( ) ( )x x y= − + −cos sinβ θ β θ . If substituting for the new axis in the slip surface of Eq.

48, the resulting slip surface is similar to what has been obtained in Eq. 19, only with

different constant values.

Analysis 3

When repeating analysis 2 with cohesion the Eq. 44 become:

[ ] [ ]dscdWdE )tan(sincossin)tan(cos φαααθφαθ −+−+−= ………………… (63)

Or

( )∫

−+

−−+−

+

−+

−=0

1 tan1

tan)tan(1sincos

tan1

tan44

u

udx

u

uukcu

u

ukE

&

&&

&

&

φζ

θβθθφζ

γ

………………………………….. (64)

And the variational equation becomes:

33

0)(2)( 63522

41 =+++−+ auauauauaua && …………………………..…………… (65)

Where:

( )( )

( )( )

( )hca

hca

hca

ka

ka

ka

−−=

−−−−=

−−−−−

−=

−=

+=

6

5

24

43

42

241

tan)tan(tantantan

)tan(tantantantantan

tantancos

tantantancos

tantantancos

φθβφζφθβφθφζφ

ζθθγ

ζθφθγ

φφθθγ

…………………………….. (66)

Thus the maximum E occurs at

63

41

2

63

52

63

52

aua

aua

aua

aua

aua

aua

du

dx

++

−

+

+−

+

+= ……………………………………….. (67)

This clearly Eq. 67 will not produce a line slip surface for c not equal zero and h = 0 also

it can be shown that h cannot be zero for maximum or minimum E. Validation of the

solution at h = 0 can be checked numerically against US Army Corps Publication EM

1110-2-2502 Equation 3-25 as a different mathematical form of the same solution.

34

Analysis 4

For a sloped smooth wall with a sloped back fill with c = 0 (as in Fig. 9(a)) the slip

surface for a passive condition can be derived similarly for shear due to the ramp. It is

assumed that

xdxKx

Kdx

dTi tan

cos

)(tantancos

cos

)tan(

2

1 2

0

2

0 φθ

θβφθ

θθβ −

=

−= ………….… (68)

Where the shear is assumed to occur from an average at rest pressure

)sin1(06.10 φ−=K on the ramp subtracting from the weight dW. From the force yields

( )[ ]θφαθ sintancos ++= dWdE ............................................................................. (69)

Thus,

( )[ ][ ]∫ +++=0

0sintancos

x

dxAxyE θφαθγ ............................................................. (70)

where ( ) dxxKydxAxydW

−−−+=+= φ

θθβ

θβ tancos

)(tantan)(

2

0 . Where

( ) φθ

θβθβ tan

cos

)(tantan

2

0

−−−= KA . Eq. 70 can be rewritten as

udxu

ukE

x

∫

′+

+′−=

0

04 costan1

tansin θ

φζ

θγ .................................................................... (71)

35

where φtan14 Ak −= ,

−−−−−= − φ

θθβ

θβφζ tancos

)(tan)tan(tan

2

01 K ,

[ ] 4/ kAxyu += , and [ ] 4/ kAyu +′=′ . From variational method, and repeating the same

analysis as in analysis 1, it yields

hu

uk

du

dx

+−−= 1tanφ ............................................................................................... (72)

and the slip surface

khuhuuh

huukux +

+++−+−−= 22ln2

tan 221φ ........................................ (73)

where ( )[ ] ( )ζθφζφ tantan/tantantan11 −−=k , and k is the constant of integration.

Substituting Eq. 72 in Eq. 71 and rewriting yields

∫

+

++=

0

1 2234

2u

uudu

huu

hukkkE γ .............................................................................. (74)

where

u y k0 0 4= / ,

[ ] 4011 / kAxyu += ,

( )( )ζθφζφθ tantantantan1tancos2 −−=k , and

( )φζθφθ tantan21costansin3 −+=k .

36

For E Eq. 51 applies.

Note: E is minimum at h = 0, giving the slip surface to be a line. Note: as the slip surface

approach the top surface the shear of Eq. 68 at y < 0 can be reduced. If h = 0 is used in

Eq. 51 it yields:

2

0

4

2

2

0

4

3

2y

k

ky

k

kE γγ += ………………………………………………………. (75)

In here 4

00

k

yu = and 01 =u , comparing with log-spiral in Table 1 at 0>= βφ and

0=θ yields:

Table 1 shows a maximum %5± difference between log-spiral and Eq. 75.

TABLE 1 – Passive Pressure Coefficient Kp for Sloped Backfill 0>= βφ and 0=θ

φ (degrees) Kp (Coluomb) Kp (log-spiral) Kp (Eq. 75)

10 1.704 1.642 1.697

15 2.321 2.170 2.284

20 3.312 3.119 3.172

25 5.074 4.822 4.600

30 8.743 7.472 7.107

35 18.82 12.67 12.14

40 70.92 23.58 24.84

37

Analysis 5

Analysis 5 can be done similar to analysis 4. Where the ramp shear is

−+

−= xc

xK

dx

dTi )tan(tancos

cos

)tan(

2

12

0 θβφθθθβ

…………………… (76)

Or

dxcxdxKTi )tan( tancos

)(tan 2

0 θβφθ

θβ−−

−−= …………. Passive ………… (77)

Surcharge

If there is a uniform surcharge q on top of the wall, it can be replaced by soil such that the

soil thickness above y is y qs = / γ . Thus y becomes y ys+ in the integral equations. By

making a change of variable y y yt s= + , the results will be similar and the equations

easily modifiable.

Wall Pressure

The location of the Coulomb force is at (2/3)y0 from the top of wall for triangular

pressure. This is a reasonable assumption since Ka is independent of y0, yielding dE/dy0 =

γKay0. However, this method is not applicable for examples 1, 2 and 3. A different

approach is necessary in order to find the pressure. This can be done by moving down the

wall from the top at incremental distances y + ∆y, and find the potential slip surfaces, and

forces. Thus, a table of Ej and yj can be created. The stresses at a distance yj can be taken

as (Ej − Ej-1) / (yj − yj-1). This will give the same result as in a Coulomb condition. For

38

example 1, 2, and 3, this method will take into account the Coulomb conditions at the top

of the wall. One needs to examine the wall boundaries and movements before using the

method. The following considerations need to be examined: (1) The potential slip

surfaces above y0 assumes the friction is almost fully mobilized. (2) The friction on the

wall may vary. It may not be constant throughout. (3) No abrupt changes in deflection

(the wall is continuous). In assumption (1), the friction on the bottom of the slice is φ φ'≤ .

However, if the wall movements indicate the entire wedge is either active or passive, then

φ' ≅ φ. φ' ≠ φ because the soil above and below yj are moving together resulting in the

apparent slip surface on the bottom of the wall. However, they must be close. In

consideration (2), even on a Coulomb wedge, if the friction of the wall varies it will

produce non-linear pressure.

Irregular Backfill

If the wall does not have a level or sloped backfill but an irregular backfill, the analysis

will require a computer program algorithm. One possibility is to use Fourier Series

converted to a Taylor Series on the top surface and proceed to solve the Euler equation.

Conclusion

Variational method has been used to determine active and passive forces for a smooth

wall with a cohesive and cohesionless soil. The methods are classical, conventional, and

only practical assumptions were used. The resulting slip surface shows that the extremum

of the force for a level backfill occurs when the slip surface is the Coulomb line for

cohesion and cohesionless soil. Additionally, in order to have the Coulomb failure

39

surface, the internal shear in the Bishop slices is required to be zero. For a sloped backfill

the resulting slip surface shows that the extremum of the force occurs when the slip

surface is a curve for cohesion soil and also when the internal shear in the Bishop slices is

not zero. When adding the internal shear due to a sloped backfill for a cohesiveless

material on a vertical wall the passive pressure is in good agreement with log spiral

method; however, the slip surface is still a line.

For special cases where the slip surface is dictated by physical conditions and must pass

through a point, a line or a curve, the forces and the slip surface can be obtained for both

active and passive conditions. Also, a method of calculating the pressures on the wall is

given. It can be seen that this method is adequate and will always give the derived slip

surface. It has been noted in the examples given that the derived slip surface has not

produced a significant difference over using a straight line. Although these differences are

of desirable accuracies, having the correct slip surface is important when considering the

influence of neighboring structures, activities, or discontinuities. If the engineer decides

to use a conservative pressure by ignoring the neighboring structure and use a slip surface

such that the neighboring structures do not exist, then the analysis given in this paper

should guide the engineer on the differences in the pressures and give him the necessary

comfort factor.

40

Appendix I

0tan

tan21

212 =

+

−−′−′

hy

hyxx

φφ

Solving the quadratic equation yields

1

212 tan

tantanhy

hyx

+

−+±=′

φφφ

or

1cos

1tan

hy

yx

+±=′

φφ

or

hc

y

cy

x

++

+±=′

φγ

φγφ

φ

tan

tan

cos

1tan

41

Appendix II.-References

1. Baker, R. and Garber, M. (1978). "Theoretical Analysis of the Stability of Slopes,"

Géotechnique, 28, No. 4, London, England, pp. 395-411.

2. Bishop, A. W. (1955). "The Use of Slip Circle in the Stability Analysis Analysis of

Slopes," Géotechnique, London, England, Vol. 5, No. 1, pp. 7-18.

3. Broms, B. B. (1968). "Swedish Tie-Back System for Sheet Pile Walls," Proc. 3rd

Budapest Conf. Soil Mech. Found. Eng., Budapest, pp. 391-403.

4. Coulomb, Charles Augustin (1776). "Essai sur une application des règles de maximis

et minimis à quelques problèmes de statique relatifs à l'architecture," Mem. Div.

Savants, Acad. Sci., Paris, Vol. 7.

5. Kranz, E. (1953). "Über die Verankerung von Spundwänden," 2 Auflage, Mitteilungen

aus dem Gebiete des Wasserbaues und der Baugrundforschung, Heft 11, Berlin: With,

Ernst and Sohn.

6. Leshchinsky, D. (1990). "Slope Stability Analysis: Generalized Approch," Journal of

Geotechnical Engineering, ASCE, Vol. 116, No. 5, pp. 851-867.

7. Ranke, A., and Ostermayer, H. (1968)." Beitrag zur Stabilitäts-Untersuchung mehrfach

verankerter Baugrubenumschliessungen" (A Contribution to the Stability Calculation

on Multiple Tie-Back Walls), Die Bautechnick, Vol. 45, No. 10, pp. 341-349.

8. Rankine, W. J. M. (1857). "On the stability of loose earth," Trans. Royal Soc., London,

147.

9. Rosenfarb, J. L., and W. F. Chen (1972). "Limit Analysis Solutions of Earth Pressure

Problems," Fritz Laboratory Roport No. 355.14, Lehigh University, pp. 53.

42

10. Shields, D. H., and Tolunay A. Z. (1973). "Passive Pressure Coefficients by Method of

Slices," Journal of Soil Mechanics and Foundations Division, ASCE, Vol. 99, No.

SM12, pp. 1043-1053.

11. Terzaghi, Karl (1941). "General Wedge Theory of Earth Pressures," Trans. Am. Soc.

Civil Eng. No.106, Vol. 67, pp. 68-80

12. Terzaghi, Karl (1943). Theoretical Soil Mechanics, Wiley and Sons, New York.

13. Terzaghi, Karl, and Ralph B. Peck (1967). Soil Mechanics in Engineering Practice,

2nd ed., John Wiley, New York.

14. U.S. Department of Commerce, National Technical Information Service, PB-257 212,

(1976). "Lateral Support Systems and Underpinning Vol. II and III : Construction

Methods" D. T. Golderg, et al., Goldber-Zoino and Asso., Inc., Newton Falls, Mass.,

pp. 180-196 in Vol. I and pp. 166-185 in Vol. II.

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Engineering, Dover Publications, Inc., New York, pp. 22-25, and pp. 36-40.

43

Appendix III.- Notation

The following symbols are used in this paper:

α = angle of the failure wedge, or of failure a slice, with the horizontal;

A = constant;

A = slope of line equation ( y = ax + b );

β = angle from the horizontal for ramp of soil on top of wall;

B = the y-axis intercept of line equation ( y = ax + b );

C = cohesion;

∆y = incremental vertical distance;

E = horizontal active force on the wall to maintain equilibrium;

Ei = horizontal active force on a slice to maintain equilibrium;

Ej = horizontal active force on wall at distance yj;

φ = angle of internal friction of soil;

φ' = immobile angle of friction soil;

f(y) = mathematical function in calculating the weight of soil in the slip surface;

γ = soil unit weight;

C = cohesion

g(x, u) = curve function where the slip surface must pass through in (x, u) coordinates;

g0(x, y) = curve function where the slip surface must pass through in (x, y) coordinates;

H = mathematical coefficient in the slip surface equations related to h0;

h0 = mathematical coefficient in the slip surface equations;

44

I = integer counter;

J = integer counter;

K0 = at rest earth pressure coefficient;

K = constant of integration;

k1 to k6 = mathematical coefficient for representations slip surface and active force Eq.'s;

Ka = active earth pressure coefficient;

Kp = passive earth pressure coefficient;

λ = dimensionless coefficient;

N = integer counter;

Pa = horizontal active force on a tieback wall to maintain equilibrium;

pa = horizontal active force above center of tieback grout length;

Q = reactive force on bottom of failure wedge or slice to maintain equilibrium;

Q = uniform surcharge pressure on top of wall;

q(y) = mathematical horizontal pressure function;

ℜ = calculus of variation function of mixed variables representing the integrand;

R(u1) = a non dimensional function = ( / ) / ( / )∂ ∂ ∂ ∂g x g u u u x x at and = =1 0 ;

R = the radius of a circle;

θ = angle from the vertical for slant wall;

T = Tmax;

Tdesign = design tieback tension force;

Ti = vertical shearing force on a slice to maintain equilibrium;

Tmax = maximum tieback tension force to failure;

45

U = new variable of distance as function of x and y;

u0 = y0;

u1 = new variable of distance as function of x0 and y1;

W = vertical force from soil weight;

ξ = tieback angle from horizontal;

X = coordinate x-axis;

x = rotated coordinate of x-axis;

x0 = x-coordinate at the end of the slip surface on top of the wall or in the soil;

xc = the x-coordinate of the center of a circle;

ψ = directional angle from the vertical for the reactive force Q;

Y = coordinate y-axis;

y = rotated coordinate of y-axis;

y0 = height of wall;

y1 = y-coordinate point on the end of the slip surface in the soil;

y2 = distance from top of the wall to tieback at face of the wall;

y3 = distance from top of the wall to Coulomb wedge;

yc = the y-coordinate of the center of a circle;

yj = the y + ∆y incremental distance on the wall for active force Ej;

ys = equivalent soil height to produce surcharge pressure q;

yt = y + ys;

ζ = φ − β + θ; and

Z = dimensionless coefficient.